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Chapter 9: Problem 44

A particle is acted simultaneously by mutually perpendicular SHM \(x=a \cos \omega t\) and \(y=a \sin \omega t\) The trajectory of motion of the particle will be. (A) An ellipse (B) A parabola (C) A circle (D) A straight line

Short Answer

Expert verified
The trajectory of the particle's motion is (C) A circle, given by the equation \(x^2 + y^2 = a^2\).

Step by step solution

01

Recognize the Given Information

We are given the following information: - Mutually perpendicular simple harmonic motion (SHM) - Equations of motion: \(x = a\cos{\omega t}\) and \(y = a\sin{\omega t}\)
02

Eliminate the Time Variable

In order to find the trajectory of the particle, we need to eliminate the time variable \(t\) from the given equations of motion. We can do this by squaring both equations and then adding them together: \[(x)^2 = (a\cos{\omega t})^2 \Rightarrow x^2 = a^2\cos^2{\omega t}\] \[(y)^2 = (a\sin{\omega t})^2 \Rightarrow y^2 = a^2\sin^2{\omega t}\] Now we can add both equations: \[x^2 + y^2 = a^2\cos^2{\omega t} + a^2\sin^2{\omega t}\]
03

Simplify the Expression

We can simplify the expression by factoring out the common term \(a^2\): \[x^2 + y^2 = a^2(\cos^2{\omega t} + \sin^2{\omega t})\] Since we know that \(\cos^2{\omega t} + \sin^2{\omega t} = 1\) (by the Pythagorean identity), further simplification yields: \[x^2 + y^2 = a^2\]
04

Identify the Trajectory

The relation between \(x\) and \(y\) is \(x^2+y^2=a^2\), which is the standard equation for a circle centered at the origin with a radius of \(a\). Therefore, the trajectory is: (C) A circle

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

SHM Equations
Simple Harmonic Motion (SHM) is crucial in understanding various physical systems and is characterized by its specific equations that describe the motion of objects oscillating back and forth. In the context of SHM, two paramount equations are typically used to describe the oscillation along perpendicular axes:\begin{align*} x &= a \times \text{cos}(\text{\(\backslash\)omega\(} t) \ y &= a \times \text{sin}(\text{\)\backslash\(omega\)} t)\text{\end{align*}}Where, in these equations:
  • x represents the displacement along the horizontal axis
  • y represents the displacement along the vertical axis
  • a symbolizes the amplitude of the motion, which is the maximum displacement from the equilibrium position
  • \(\backslash\)omega indicates the angular frequency, translating to how fast the oscillation occurs
  • t stands for time, the independent variable in both equations
The expressions provide a description of how the position of a particle in SHM changes over time along each axis.
Trajectory of SHM
Discussing the trajectory of an object in SHM leads us to visualize the path traced by the object during its motion. In the given problem, we're examining the trajectory of a particle experiencing SHM along two mutually perpendicular axes鈥攈orizontal and vertical. The provided equations for SHM represent sinusoidal functions describing the oscillations in each direction. To determine the trajectory:\begin{align*} x &= a \times \text{cos}(\text{\(\backslash\)omega\(} t) \ y &= a \times \text{sin}(\text{\)\backslash\(omega\)} t) \text{\end{align*}}The equations are connected through the common factors of amplitude a, angular frequency \(\backslash\)omega, and time t. Theoretically, by eliminating the t variable, we can find a relation purely between x and y, which gives us the trajectory of the particle in the form of a curve or geometric shape. As shown in the provided step-by-step solution, this manipulation leads to an equation representative of a circle with radius a. This showcases how the superposition of two orthogonal SHM movements can create a circular motion.
Pythagorean Identity in SHM
The Pythagorean identity is a fundamental aspect of trigonometry and is closely linked with SHM when analyzing motions in two dimensions. The identity states that for any angle \(\backslash\)theta:\begin{align*} \text{cos}^2(\text{\(\backslash\)theta\(}) + \text{sin}^2(\text{\)\backslash\(theta\)}) = 1 \text{\end{align*}}In the context of SHM, when a particle exhibits simultaneous perpendicular motions described by x = a \text{cos}(\text{\(\backslash\)omega\(} t) and y = a \text{sin}(\text{\)\backslash\(omega\)} t), the time variable t can be eliminated to yield:
\begin{align*} x^2 + y^2 = a^2(\text{cos}^2(\text{\(\backslash\)omega\(} t) + \text{sin}^2(\text{\)\backslash\(omega\)} t))\text{\end{align*}}Applying the Pythagorean identity simplifies this to x^2 + y^2 = a^2, which is a direct representation of a circle's equation in Cartesian coordinates. Understanding this connection between SHM equations and the Pythagorean identity allows students to easily visualize and grasp the circular trajectory followed by the particle undergoing such motions.

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Most popular questions from this chapter

A sound wave of wavelength \(\lambda\) travels towards the right horizontally with a velocity \(v\). It strikes and reflects from a vertical plane surface, traveling at a speed \(v\) towards the left. The number of positive crests striking in a time interval of three seconds on the wall is (A) \(3(V+v) / \lambda\) (B) \(3(V-v) / \lambda\) (C) \((V+v) / 3 \lambda\) (D) \((V-v) / 3 \lambda\)

The amplitude of a wave disturbance propagating in the positive \(y\)-direction is given by \(y=\frac{1}{1+x^{2}}\) at \(t=\) 0 and \(y=\frac{1}{\left[1+(x-1)^{2}\right]}\) at \(t=2\) second, where \(x\) and \(y\) are in \(m\). If the shape of the wave disturbance does not change during the propagation, what is the velocity of the wave? (A) \(1 \mathrm{~m} / \mathrm{s}\) (B) \(1.5 \mathrm{~m} / \mathrm{s}\) (C) \(0.5 \mathrm{~m} / \mathrm{s}\) (D) \(2 \mathrm{~m} / \mathrm{s}\)

A wave is represented by the equation: \(y=0.1 \sin (100 \pi t-k x) .\) If wave velocity is \(100 \mathrm{~m} / \mathrm{s}\), its wave number is equal to (A) \(1 \mathrm{~m}^{-1}\) (B) \(2 \mathrm{~m}^{-1}\) (C) \(\pi \mathrm{m}^{-1}\) (D) \(2 \pi \mathrm{m}^{-1}\)

A transverse wave is described by the equation \(y=y_{0} \sin 2 \pi\left(f t-\frac{x}{\lambda}\right) .\) The maximum particle veloc- ity is equal to four times the wave velocity if (A) \(\lambda=\frac{\pi y_{0}}{4}\) (B) \(\lambda=\frac{\pi y_{0}}{2}\) (C) \(\lambda=\pi y_{0}\) (D) \(\lambda=2 \pi y_{0}\)

A massless \(\operatorname{rod} A B\) of length \(L\) is hung from two identical wires of equal length. A block of mass \(m\) is attached at point \(O\) on the rod as shown in Fig. \(9.28\); the value of \(A O\) so that a tuning fork excites the wire on the left in its fundamental tone and the wire on the right in its second harmonic is (A) \(\frac{4 L}{5}\) (B) \(\frac{L}{4}\) (C) \(\frac{3 L}{4}\) (D) \(\frac{L}{5}\)
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